Illinois State University Mathematics Department
MAT 409: Topics in Algebra and Combinatorics for K8 Teachers 











Chapter 3 Section
2
Notice, though, that, based on restriction (iii),
some of these are duplicates, such as the two
highlighted above [BAC and CAB]. We need to
account for these duplicates. They are highlighted
above using cell background color. Thus, we need to
remove 6, leaving us with 10 potential flags when we
fix one color for the center stripe. Note that we do
not simply divide 16 by 2, because the diagonal
entries (first and last stripe same color) do not
have a duplicate. We can repeat this with each of the four remaining
centerstripe colors. This gives us 10 * 5 = 50
possible flags under the restrictions and conditions
stated in the problem.
Because each case has no overlap
with other cases, we add to determine the total
number of committees that are possible: C(20,3) *
C(20,5) + C(20,4) * C(20,4) + C(20,5) * C(20,3) =
2[C(20,3) * C(20,5)] + [C(20,4)]^{2}.
(b) Here, it
is more efficient to determine the total number of
committees possible and remove those with no
faculty: C(40,8)  [C(20,0) * C(20,8)]. 